Suppose $a$ and $b$ are positive integers and $\dfrac{a}{4}+\dfrac{b}{3}=\dfrac{13}{12}$. What is the value of $a^2+b^2$?
Answer: Multiply both sides by 12 to rewrite the equation as $3a+4b=13$. Trying $a=1$, $2$, $3$, and $4$, we find that the only positive integer value of $a$ for which $13-3a$ is a positive multiple of 4 is $a=3$. Substituting $a=3$ into $3a+4b=13$, we find $b=1$. So, $a^2+b^2=3^2+1^2=\boxed{10}$.